Problem: Solve the equation. $\dfrac13 + a = \dfrac54$ $a=$
Let's subtract to get $a$ by itself. $\begin{aligned} \dfrac13 +a &= \dfrac54 \\ \\ \dfrac13 +a {-\dfrac13}&= \dfrac54{-\dfrac13}~~~~{\text{subtract }\dfrac13} \text{ from each side to get } a \text{ by itself }\\ \\ \cancel{ \dfrac13} +a {{-}\cancel{{\dfrac13}}}&= \dfrac54{-\dfrac13}\\ \\ a &= \dfrac54{-\dfrac13}\end{aligned}$ $\begin{aligned} \dfrac54-\dfrac13 &=\dfrac{5\times3}{4\times3}-\dfrac{1\times4}{3\times4}\\\\ &= \dfrac{15}{12}-\dfrac{4}{12} \\\\ &= \dfrac{11}{12} \end{aligned}$ The answer: $a={\dfrac{11}{12}} $ Let's check to make sure. $\begin{aligned} \dfrac13 +a &= \dfrac54 \\\\ \dfrac13 +{\dfrac{11}{12}} &\stackrel{?}{=} \dfrac54 \\\\ \dfrac{4}{12} +{\dfrac{11}{12}} &\stackrel{?}{=} \dfrac54 \\\\ \dfrac{15}{12} &\stackrel{?}{=} \dfrac54 \\\\ \dfrac54 &= \dfrac54 ~~~~~~~~~~\text{Yes!} \end{aligned}$